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An $n$-qubit stabilizer state is the unique simultaneous $+1$ eigenstate of an abelian group $G$ of Pauli operators having $2^n$ elements and not containing $-I$. The group $G$ is the state's stabilizer group; equivalently, it is generated by $n$ independent commuting Hermitian Pauli operators.

Irreducible degrees divide the order of a finite $2$-group and satisfyThe number of linear characters is $|G/G^{\prime }|$. For a nontrivial finite $2$-group this is at least $4$ unless the group is abelian, and every nonlinear degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, thenThus the only degree collections arewith respectively $r=16,10,7$.

For $1^{16}$ take the cyclic group $G=C_{16}=⟨g⟩$. Every conjugacy class is a singleton, and its character table iswhere ${\zeta }_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ and
$C_2=⟨z⟩$. On the five classesthe four linear characters of $D_8$, indexed by
$\epsilon ,\delta \in \{\pm 1\}$, and its degree-two character areThe ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists ofwhere $\chi$ runs over the above five rows and
$\eta \in \{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the dihedral groupOrder its seven classes asThe four linear rows areThe three degree-two rows, for $k=1,2,3$, areThe row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.

Embed $E$ as a smooth plane cubic with $O_E$ an inflection point. A line through $P,Q$, using the tangent when $P=Q$, has a third intersection $S$ counted with multiplicity. The chord-and-tangent group law defines $P+Q=-S$, where $-S$ is the third point on the line through $S$ and $O_E$.

The divisor cut out by the first line isand the line through $S$ and $O_E$ gives $(S)+(-S)-2(O_E)$. Their quotient therefore showsUnder the bijection from part (a), the chord-and-tangent operation is exactly addition in the abelian group ${\mathrm{Pic}}^0(E)$. It is consequently associative and commutative, has identity $O_E$, and has the geometrically defined point $-P$ as inverse. Thus it makes $E$ an abelian group.

Let $E$ be a smooth plane cubic whose identity $O$ is an inflection point. A line through $P$ and $Q$, using the tangent when $P=Q$, has a third intersection $R$ counted with multiplicity. The chord-and-tangent group law defines $P+Q$ by drawing the line through $R$ and $O$ and taking its third intersection.

The clean verification of the group axioms uses divisors. The line at infinity meets a Weierstrass cubic in $3(O)$, so three collinear points $P,Q,R$ satisfyConsequently the mapsends the chord-and-tangent construction to addition of divisor classes. The principal divisor criterion on an elliptic curve shows that this map is bijective. Associativity and commutativity therefore follow from the abelian group law on ${\mathrm{Pic}}^0(E)$. The tangent convention handles repeated intersections, $O$ represents the zero class, and the third point on the line through $P$ and $O$ represents the inverse of $P$. Hence all group axioms hold.

The Plünnecke-Ruzsa inequality says that if $A,B$ are finite nonempty subsets of an abelian group andthen for all nonnegative integers $m,n$,

Choose a nonempty $X\subseteq A$ minimizingthen $K^{\prime }\le K$. We first prove the Petridis minimal-growth lemmafor every finite $C$, by induction on $|C|$. Remove $c\in C$, write $C^{\prime }=C\setminus \{c\}$, and letThe new points contributed to $X+C$ by $X+c$ are exactly $|X\setminus X^{\prime }|$. Moreover, $X^{\prime }+B+c\subseteq (X+B+c)\cap (X+B+C^{\prime })$, soThe induction hypothesis, the identity $|X+B|=K^{\prime }|X|$, and minimality, which gives $|X^{\prime }+B|\ge K^{\prime }|X^{\prime }|$, yieldIteration with $C=(r-1)B$ gives

Finally, the Ruzsa triangle inequality givesas required.

The Schur multiplier of a group $G$ isthe second group homology group with trivial integral coefficients. If $G\cong F/R$ is a free presentation, Hopf's formula states that

Write $I_F=\ker (\mathbb{Z}F\to \mathbb{Z})$ for the augmentation ideal. The presentation relation sequence iswhere $r[R,R]\mapsto 1\otimes (r-1)$. If $F$ is free on a set $S$, then $I_F$ is free as a left $\mathbb{Z}F$-module on the elements $s-1$, so the two modules immediately preceding $\mathbb{Z}$ are free $\mathbb{Z}G$-modules. Resolving the relation module $R/[R,R]$ by free modules and splicing produces a free resolution of $\mathbb{Z}$.

Apply the right-exact functor $\mathbb{Z}{\otimes }_{\mathbb{Z}G}-$ to this partial resolution. Its degree-two homology is the kernel ofThe coinvariant module on the left is $R/[F,R]$. On the right, the map $f-1\mapsto f[F,F]$ identifies the coinvariants with the abelianization $F/[F,F]$. The displayed map is induced by the inclusion $R\hookrightarrow F$, so its kernel isThis proves Hopf's formula.

For an abelian group $A$, the Schur multiplier of an abelian group is $M(A)\cong {\bigwedge }^{2}A$. One way to see the direct-sum rule is the degree-two Künneth theorem:A cyclic group has zero second integral group homology, whileConsequently

The integers are closed under addition, addition is associative, zero is an identity, and $-n$ is the additive inverse of $n$. Thus $(\mathbb{Z},+)$ is an abelian group.

The statement that limits of shape $\mathcal{I}$ commute with colimits of shape $\mathcal{J}$ means that for every $D:\mathcal{I}\times \mathcal{J}\to \mathcal{C}$, the canonical commutation of limits and colimits mapis an isomorphism whenever the iterated limits and colimits exist.

A filtered category is a nonempty category in which every finite diagram has a cocone. Equivalently, any two objects map to a common object and any parallel pair becomes equal after postcomposition. A weakly filtered category requires cocones only for finite connected diagrams; equivalently, each connected component is filtered.

Write a weakly filtered category as the disjoint union ${\coprod }_s{\mathcal{J}}_s$ of its filtered connected components. Its colimit is the coproduct of the filtered colimits over the ${\mathcal{J}}_s$. In sets, a connected finite limit commutes with coproducts: connectedness forces all coordinates of a compatible tuple to lie in the same coproduct summand. By the assumed theorem, the filtered colimit over each ${\mathcal{J}}_s$ commutes with every finite limit. Applying these two facts successively proves that weakly filtered colimits commute with connected finite limits in $\mathbf{Set}$.

The forgetful functor from abelian group to sets creates finite limits and filtered colimits and reflects isomorphisms. The comparison map for a filtered colimit and a finite limit therefore becomes the corresponding isomorphism of sets, so filtered colimits commute with finite limits in $\mathbf{AbGp}$.

The dual claim fails because inverse limit need not preserve epimorphisms. Take the inverse systemswith identity bonding maps on $A_n$, reduction maps on $B_n$, and levelwise epimorphisms $q_n:A_n\to B_n$. Thenand the induced map $\mathbb{Z}\to {\mathbb{Z}}_2$ is not surjective. Since an epimorphism in abelian groups is a finite-colimit cokernel, cofiltered limits do not commute with finite colimits in $\mathbf{AbGp}$.

The Entropic Balog-Szemerédi-Gowers theorem states that for finitely supported random variables $A,B$ in an abelian group,where the left-hand side is the Simultaneous conditional entropic Ruzsa distance.

Put $S=A+B$ and take two copies $(A_1,B_1)$ and $(A_2,B_2)$ that are conditionally independent given $S$. Thus both sums equal $S$, and, given $S$, $A_1$ and $B_2$ are independent with the required conditional marginals. By conditioning reduces entropy,The two conditional entropies are equal, since either $A$ or $B$ together with $S$ determines the other, and the chain rule for information entropy gives

Set $D=A_1-B_2$. By entropy submodularity,The first term is $H(A_1,B_2)\le H(A)+H(B)$ by subadditivity of information entropy. Since $A_1+B_1=A_2+B_2$, we also have $D=A_2-B_1$, so the second term is at most $H(A)+H(B)$. Finally $(D,A_1,B_1)$ determines all four copied variables, and conditional independence givesAs $H(A,B)=H(A)+H(B)-I(A;B)$, these estimates implySubtracting the common conditional-entropy term proves the stated Entropic Balog-Szemerédi-Gowers theorem.

The Dold–Kan correspondence says thatis an equivalence from simplicial abelian groups to nonnegatively graded chain complexes of abelian groups. The restriction of the right adjoint $K$ to ${\mathbf{Ch}}_{\ge 0}(\mathbf{Ab})$ is a quasi-inverse: both the unit $C_{\ast }\to N_{\ast }K(C_{\ast })$ and counit $K(N_{\ast }A)\to A$ are natural isomorphisms.

Suppose $\sigma \ne e$. Choose a symbol $i$ moved by $\sigma$ and write $j=\sigma (i)\ne i$. Since $n\ge 3$, choose a third symbol $k$ distinct from $i,j$. Let $\tau =(jk)$, a transposition that fixes $i$. ThenThus ${\tau }^{-1}\sigma \tau \ne \sigma$, proving that the conjugacy class of any nonidentity permutation has more than one element. By part (a)(iii), no such permutation lies in the center of a group. The identity always does, soThe restriction on $n$ is necessary: $S_2$ is an abelian group and equals its own center.

The hypothesis implies $x^{-1}=x$ for every $x\in G$, including $e$. For $a,b\in G$,Thus $G$ is an abelian group. This is the group of exponent two is abelian argument.

The converse is false: the cyclic group $C_3$ is an abelian group, but its two nonidentity elements have order of a group element three.

If $G=⟨g⟩$, any two elements are $g^a,g^b$, andThus every cyclic group is an abelian group.

For the converse, the Klein four-group $C_2\times C_2$ is an abelian group, with componentwise multiplication. Each nonidentity element has order of a group element two, so none generates its four elements. It is therefore not a cyclic group.

Fix $x\in G$. The integer powers of a group element satisfyFor nonnegative integers these rules follow by associativity and counting factors; negative integers are handled by $x^{-r}=(x^{-1}{)}^r$, and opposite signs by cancellation of adjacent inverse elements. All factors are powers of the same element, so no abelian group assumption is involved.

It follows that $z\mapsto \rho (z,x)=x^z$ is a group homomorphism from the additive group of integers to $G$, andThe homomorphism uses ordinary addition on $\mathbb{Z}$; composition of the power functions corresponds to ordinary multiplication of exponents.

Suppose the quotient group $G/A$ is a cyclic group, generated by $gA$. For any $x,y\in G$, there are integers $r,s$ and $a,b\in A$ such thatSince $a,b$ lie in the center of a group, they commute with every element, in particular with each other and with the integer powers of a group element $g$. ThereforeAs $x,y$ were arbitrary, $G$ is an abelian group. This is the cyclic quotient by the center.

A vector space over a field $\mathbb{F}$ is a set $V$ with vector addition $V\times V\to V$ and scalar multiplication $\mathbb{F}\times V\to V$ satisfying the following vector space axioms. Under vector addition, $V$ is an abelian group: addition is associative and commutative, there is a zero vector $0_V$, and each $v$ has an additive inverse $-v$. For all $a,b\in \mathbb{F}$ and $u,v\in V$,The operations are defined for every indicated pair and take their values in $V$.

If $G$ is an abelian group, its elements commute, so for all $x,y\in G$,Thus $\boxed{T\text{ is a group homomorphism}}$. Also $T(e)=e^3=e$, as required.

If $T$ is a surjective group homomorphism, then $G=T(G)$, which is an abelian group by the preceding part.

If $T$ is an injective group homomorphism, then for all $a,b\in G$,because $T(a),T(b)$ lie in the abelian group $T(G)$. Injectivity gives $ab=ba$. Hence $\boxed{G\text{ is abelian in either case}}$.

Yes: use the non-abelian group $H$ from part (a). Every element is $I_3+N$ for a strictly upper triangular matrix $N$ with $N^3=0$. In the finite field ${\mathbb{Z}}_3$, the binomial theorem for commuting matrices givesThus $\boxed{T(h)=I_3\text{ for all }h\in H}$. This constant identity-valued map is a group homomorphism, since $T(xy)=I_3=T(x)T(y)$. Part (a)(iv) already proves that $H$ is not an abelian group. Here $T$ is neither injective nor surjective, consistently with the preceding part.

Conjugacy classes. For $a\in G$, its conjugacy class isEvery element belongs to its own conjugacy class, by taking $g=e$. To prove uniqueness, suppose $x=ga{g}^{-1}=hb{h}^{-1}$ belongs to both $\mathrm{Cl}(a)$ and $\mathrm{Cl}(b)$. ThenFor every $t\in G$, this makes $tb{t}^{-1}$ a conjugate of $a$ as well. Hence $\mathrm{Cl}(b)\subseteq \mathrm{Cl}(a)$, and reversing the argument gives the other inclusion. Two conjugacy classes that meet are therefore equal, proving that the classes partition $G$.

Abelian case. If $G$ is an abelian group, then $ga{g}^{-1}=ag{g}^{-1}=a$ for every $g$, so $\boxed{\mathrm{Cl}(a)=\{a\}}$ for each $a$.

Centre. An element has a singleton conjugacy class precisely when $gz{g}^{-1}=z$ for every $g$, equivalently $gz=zg$ for every $g$. Thus the definition in the paper agrees with the usual centre of a group:The identity lies in $Z(G)$. If $z,w\in Z(G)$, then for any $g$, $gzw=zgw=zwg$, so $zw\in Z(G)$. Also $gz=zg$ implies $z^{-1}g=g{z}^{-1}$, so $z^{-1}\in Z(G)$. This proves it is a subgroup. Any two of its elements commute because either one commutes with every element of $G$. Consequently $\boxed{Z(G)\text{ is an abelian subgroup of }G}$.

Square symmetry. Use the paper's notation $D_4$ for the eight-element dihedral group of a square. Let $r$ be a quarter-turn and $s$ a reflection, so $r^4=s^2=e$ and $srs=r^{-1}$. A rotation $r^j$ commutes with $s$ exactly when $r^j=r^{-j}$, or $2j=0$ modulo four. This selects $e$ and $r^2$. No reflection $r^{j}s$ commutes with $r$, becauseand $r^{j-1}\ne r^{j+1}$. Therefore